NCERT Solutions Class 9th Maths Chapter – 1 Number System Exercise – 1.3

Question 1. Write the following in decimal form and say what kind of decimal expansion each has:

(i) 36/100

Solution

= 0.36 (Terminating)

(ii) 1/11

Solution

Solution

= 4.125 (Terminating)

(iv) 3/13

Solution

(v) 2/11

Solution

(vi) 329/400

Solution

= 0.8225 (Terminating)

Question 2. You know that 1/7 = 0.142857¯. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?

Solution – We are given that 17 = 0.142857¯
2/7 = 2 x 1/7 = 2 x (0.142857¯) =0.285714¯
3/7 = 3 x 1/7 = 3 x (0.142857¯) = 0.428571¯
4/7 = 4 x 1/7 = 4 x (0.142857¯) = 0.571428¯
5/7 = 5 x 1/7 = 5 x(0.142857¯) = 0.714285¯
6/7 = 6 x 1/7 = 6 x (0.142857¯) = 0.857142¯
Thus, without actually doing the long division we can predict the decimal expansions of the given rational numbers.

Question 3. Express the following in the form p/q, where p and q are integers and q ≠ 0.

(i) 0.6¯

Solution – Let x = 0.6¯ = 0.6666… … (1)
As there is only one repeating digit,
multiplying (1) by 10 on both sides, we get
10x = 6.6666… … (2)
Subtracting (1) from (2), we get
10x – x = 6.6666… -0.6666…
⇒ 9x = 6 ⇒ x = 6/9 = 2/3
Thus, 0.6¯ = 2/3

(ii) 0.47¯

Solution – Let x = 0.47¯ = 0.4777… … (1)
As there is only one repeating digit, 
multiplying (1) by lo on both sides, we get
10x = 4.777
Subtracting (1) from (2), we get
10x – x = 4.777…… – 0.4777…….
⇒ 9x = 4.3 ⇒ x = 43/90
Thus, 0.47¯ = 43/90

(iii) 0.001¯¯

Solution – Let x = 0.001¯¯¯¯¯¯¯¯ = 0.001001… … (1)
As there are 3 repeating digits,
multiplying (1) by 1000 on both sides, we get
1000x = 1.001001 … (2)
Subtracting (1) from (2), we get
1000x – x = (1.001…) – (0.001…)
⇒ 999x = 1 ⇒ x = 1/999
Thus, 0.001¯¯¯¯ = 1/999

Question 4. Express 0.99999… in the form p/q Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Solution – Let x = 0.99999….. …. (i)
As there is only one repeating digit,
multiplying (i) by 10 on both sides, we get
10x = 9.9999 … (ii)
Subtracting (i) from (ii), we get
10x – x = (99999 ) — (0.9999 )
⇒ 9x = 9
⇒ x = 9/9
⇒ x = 1
Thus, 0.9999 = 1

Question 5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.

Solution – In 1/17, In the divisor is 17.
Since, the number of entries in the repeating block of digits is less than the divisor, then the maximum number of digits in the repeating block is 16.
Dividing 1 by 17, we have
The remainder I is the same digit from which we started the division.
∴ 1/17 = 0.0588235294117647¯
Thus, there are 16 digits in the repeating block in the decimal expansion of 1/17.
Hence, our answer is verified.

Question 6. Look at several examples of rational numbers in the form p/q (q ≠ 0). Where, p and q are integers with no common factors other that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Solution – We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion terminates. For example:
1/2 = 0. 5, denominator q = 2¹
7/8 = 0. 875, denominator q =2³
4/5 = 0. 8, denominator q = 5¹
We can observe that the terminating decimal may be obtained in the situation where the prime factorisation of the denominator of the given fractions has the power of only 2 or only 5 or both.

Question 7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Solution
√2 = 1.414213562 ………..
√3 = 1.732050808 …….
√5 = 2.23606797 …….

Question 8. Find three different irrational numbers between the rational numbers 5/7 and 9/11 .

Solution

Three irrational numbers between 0.714285¯ and 0.81¯ are
(i) 0.750750075000 …..
(ii) 0.767076700767000 ……
(iii) 0.78080078008000 ……

Question 9. Classify the following numbers as rational or irrational:

(i) √23

Solution – √23 = 4.79583152331…
√23 is an irrational number.

(ii) √225

Solution – √225 = 15 x 15 = 15²
√225 is a perfect square.
Thus, √225 is a rational number.

(iii) 0.3796

Solution – 0.3796 is a terminating decimal.
It is a rational number.

(iv) 7.478478…..

Solution – 7.478478… = 7.478¯
Since, 7.478 is a non-terminating recurring (repeating) decimal.
∴ It is a rational number.

(v) 1.101001000100001………

Solution – Since, 1.101001000100001… is a non terminating, non-repeating decimal number.
∴ It is an irrational number.